In Exercises 21–32, express the integrand as a sum of partial ...

Question

In Exercises 21–32, express the integrand as a sum of partial fractions and evaluate the integrals.

∫ (x² + x) / (x⁴ - 3x² - 4) dx

Accepted Answer

Hello. In this video, we are going to be evaluating the following integral. We are given the integral of x2 minus x divided by x 4 plus 2x2 + 1. Now, in order to approach this problem, the first thing we want to do is you want to simplify the denominator by factoring. If we factor the denominator, we can rewrite the integral as the integral of X2 minus X divided by the quantity X2 + 1 quantity 2. Now here, what we can go ahead and do is we can simplify the integrand by using partial fraction decomposition. So here in our integrand, we have X2 minus X divided by X2 + 1 quantity 2. Now here we have a repeated quadratic term in our denominator, and so what we're going to get is two different partial fractions. The first one would be defined as A X + B divided by X2 + 1, plus C X + D divided by X2 + 1 quantity squad. Now, if we were to multiply by the least common denominator, we can simplify this equation to be X2 minus X equal to the quantity AX + B multiplied by X2 + 1. Plus the quantity c x + d. Now if we were to multiply and reorder terms, we can further rewrite this equation to be x2 minus x equal. To a x cubed plus B X 2. Plus the quantity A + C multiplied by X plus the constant terms B + D. Now here we can go ahead and get a system of linear equations by setting coefficients equal to each other. For the first equation, if we set the coefficients of x cubed equal to each other, we get a's equal to 0, which is the first coefficient we're looking for. Setting the x 2 terms equal to each other gives us b is equal to 1, which is a 2nd coefficient. For the third equation, setting the x term coefficients equal to each other gives us a plus c equal to -1. And since a is equal to 0, this gives us c is equal to -1. And for the 4th equation, we get B + D is equal to 0, but since B is equal to 1, plugging this into the equation gives us D is equal to -1. So, if we were to reapply this back into our partial fractions, the partial fractions can be rewritten as 0 x plus. 1 divided by X2 + 1. Minus. X + 1 divided by the quantity X2 + 1 quantity 2. So, if we were to rewrite this into the integral, we can rewrite the original integral as the integral of 1 divided by X2 + 1 minus the integral of X + 1 divided by the quantity X2 + 1 quantity squad. We will go ahead and label this first integral as integral 1, and label the second integral as integral 2. And now what we want to go ahead and do is we want to evaluate each of the integrals independently. Starting with the first integral, again, we have the integral of 1 divided by X2 + 1 DX. Now, what we can go ahead and do here is apply. Oh, is apply our tangent inverse identity. And if we were to take the integral of this of this integran by using the tangent inverse identity, this is going to evaluate to be tangent inverse of X. So this is going to be the value of our first integral. Now, the second integral is a little bit trickier to approach. Let's go ahead and ignore the negative sign for now. But in order to approach this integral, we will first rewrite this as a sum of integrals. So we can rewrite the second integral as the integral of X divided by X2 + 1 quantity 2 plus the integral of 1 divided by X2 + 1 quantity squad. Now, if we are taking a look at the first part of the 2nd integral, we can evaluate this by applying a U substitution and allowing you to equal to X2 + 1. Then DU will equal to 2 XDX, which can further be rewritten as 1/2 DU equal to XDX. Then this first integral will simplify to be 1/2 multiplied by the integral of 1 divided by u squared, and this is going to further simplify. To be -1/2 multiplied by u to the 1st. And if we were to resubstitute x back into here, this would be rewritten as -1 divided by 2 multiplied by the quantity x 2 + 1. So this is going to be the output of the first integral. Now the second integral is a little trickier to work with. For the 2nd integral, we will need to use trigonometric substitution. Now, here we can allow X to equal to tangent theta. Then dx will equal to sequin squared theta d theta. Applying this trigonometric substitution to the integral allows us to rewrite the integral as the integral of sequent squared theta divided by tangent squared theta. Plus 1 quantity squared d theta. Now here, tangent square theta plus one can be rewritten as sequin square theta. And keep in mind that sequin square theta in the denominator is also raised to the power of 2, so the entire denominator is going to simplify to be sequin to the 4th theta. And since we have two multiples of sequin theta in both the numerator and denominator, we can eliminate two of them from both the numerator and denominator and rewrite this integral as the integral of 1 divided by sequin squared theta d theta. This can further be rewritten as the integral of cosine squared theta d theta. And now if we were to use our power reduction identity for cosine squared theta, we can further rewrite the integral to be the integral of One Plus cosine of 2 theta divided by 2. Moving the coefficient of 1/2 to the front of the integral allows us to rewrite it as 1/2 multiplied by the integral of 1 plus cosine of 2 theta d theta. Now, if we were to integrate and distribute the 1/2, we can rewrite this as 1/2 theta. Plus 1/4 multiplied by sine of 2 theta. And by applying the double angle identity of sin of 2 theta, we can further simplify this to be 1/2 theta. Plus 1/2 multiplied by sine of theta multiplied by cosine of theta. Now here I am running out of a little bit of room, so let's go ahead and clear up some of this work, but we are going to keep our results for the first integral here. So this is the current form of the 2nd part of the 2nd integral. Now, recall that earlier in the problem, we use the substitution that tangent of theta is equal to x, but we can also rewrite this as tangent of theta is equal to x divided by 1. Now we can create a right triangle to try to find the values of sine theta and cosine theta respective to the substitution. By creating the right triangle. We will mark our angle theta here, and keep in mind that tangent theta is equal to the opposite over the adjacent side. So with respect to theta, that is going to give us X on the opposite side and 1 on the adjacent side. And then by using the Pythagorean identity, we will have the square root of X2 + 1 for the hypotenuse. So now if we take a look at this right triangle, sine theta of this right triangle is going to be the opposite side over the hypotenuse. So that is going to be X divided by the square root of X2 + 1. Cosine theta is the adjacent side with respect to the hypotenuse, and that's going to be 1 divided by the square root of X2 + 1. And now to sulfur theta, we can actually use the original substitution. If we were to take the tangent inverse of both sides of this equation, theta is going to equal to tangent inverse of X. So by applying these substitutions we will have 1/2. 1/2 multiplied by tangent inverse of x. Plus 1/2 multiplied by the quantity x divided by the square root of x2 + 1 multiplied by 1 divided by the square root of x2 + 1. And this is going to simplify to be 1/2 multiplied by the tangent inverse of x. Plus X divided by 2 multiplied by x2 + 1. This is going to be the result of the 2nd part of our original integral. So just to do a recap, the first integral simplified to be tangent inverse of x, and by splitting apart the second integral, we broke it apart to be -1 divided by 2 multiplied by x2 + 1, and 1/2 tangent inverse of x plus x divided by 2 multiplied by x2 + 1. Now what we're going to do is we're going to take all three of these results and combine them together. So let's go ahead and create some room for that. Now, one thing to keep in mind is that in the original problem, or the original. In the original broken down integrals. We had a difference of integrals. So by applying the results, we will have tangent inverse of x. Minus the quantity -1 divided by 2 multiplied by x2 + 1 + 1/2 tangent inverse of X plus X divided by 2 multiplied by x2 + 1, and because we were working with a generic integral, we will add a general constant C. Now once we distribute the negative sign and combine like terms, we can further simplify the solution to be 1/2 tangent inverse of X plus 1 minus X divided by 2 multiplied by X2 + 1 plus the generic constant C. And this is going to be the solution to the given integral, and the overall solution will be option A. So I hope this video helps you in understanding on how to approach this problem, and we will go ahead and see you all in the next video.

In Exercises 21–32, express the integrand as a sum of partial fractions and evaluate the integrals.∫ (x² + x) / (x⁴ - 3x² - 4) dx

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In Exercises 21–32, express the integrand as a sum of partial fractions and evaluate the integrals.
∫ (x² + x) / (x⁴ - 3x² - 4) dx